Bias Balance Ineffective- WHY?

[jjasilli]

Why is my newly installed bias balance pot ineffective?  The bias circuit works and is adustable.   But the bias balance pot has almost zero effect -- less than one volt.  How can this be?  Schematic attached.

[sluckey]

But the bias balance pot has almost zero effect -- less than one volt.  How can this be?

There is no current flowing thru your balance pot or the 100KΩ resistors. No current flow = no voltage dropped = no change in bias voltage applied to the grids.

[jjasilli]

Bingo, thanks!  I checked a schematic of another amp and now see that the outer lugs of that bias balance pot have shunt resistors to ground.  That will get current flowing through the bias balance pot.  But I think those shunt resistors will also interact with the grid leak resistors to the power tubes.  This is getting complicated!

Altenative:  a separate pot for ea power tube?  Any other suggestions?

[Geezer]

a separate pot for ea power tube?

That was going to be my suggestion.......

[Colas LeGrippa]

hook up the 10k pot as a variable resistor: use only 2 lugs, the third lug going to the 27K to ground is bad.

Colas

[sluckey]

But I think those shunt resistors will also interact with the grid leak resistors to the power tubes.

They will, but so do the resistors in the typical bias adjust circuit. However, the bias resistor values are usually small compared to the grid leak resistors and won't have much (if any) effect. Does it really matter if you have a single 220K grid leak versus a 247K grid leak made up of a string of resistors?

a separate pot for ea power tube?

If you're gonna have two pots, this just makes the most sense to me.

[FYL]

Any other suggestions?

You may try this. The text is in French but the schemo and formulas are pretty much self-explanatory.

[sluckey]

hook up the 10k pot as a variable resistor: use only 2 lugs, the third lug going to the 27K to ground is bad.

Actually, the wiper is the weak link in this particular bias circuit. If it opens (maybe just because of a speck of dust or dirt) you lose bias to the tubes. I also have concerns about this circuit (in my head), but I have never actually seen a failed bias pot. I know there must be some out there.

However, Fender has sold a bazillion amps with this identical circuit.

[jjasilli]

Colas:  Thanks for that reminder!

FYL:  Brilliant!  Ou la la!  That's the circuit I was trying to conceive of, but failed to implement. 

Sluckey & Geezer:  Because this is being retrofitted into an exisiting amp-- Traynor EL84 Bassmate -- I'll probably just go the 2 separate pots route as you suggest.  (Still the French way has a haunting elegance about it.)

[sluckey]

(Still the French way has a haunting elegance about it.)

I agree. The circuit is simple and effective. And the drawing has a very eye pleasing symmetry. However, all that symmetry dissappears when you draw the output tubes in the typical vertical position.   

[Fresh_Start]

For whatever reason, the idea of a bias adjustment pot and a bias balance pot appeals to me.  I'd been looking for a circuit with actual values for a while and came across something recently that looked good (top of attached SCH drawing). 

However, the "French style" offered up by FYL is  "hauntingly elegant" (are you writing a romance novel?). After looking at FYL’s suggestion some more, I went ahead and drew it in SCH for myself and anyone else who might like to have it handy.  (bottom portion of attached drawing)

One question:  do the two caps in Yves' circuit provide just as much filtering as the alternative.  If so, it IS more elegant.  Lower parts count too...

Second question:  If you know your raw bias voltage  (AlimNeg) and your target grid voltage (Vg), how do you calculate the value of R2 and R6? (shown as 15K in my drawing despite the 17.8K suggestion in French)  There are some parentheses missing in the formula shown for R2 & R6.

BTW our comrade FYL appears to be truly multilingual - French schematics today and a Mouser reference in German a few days ago!

In the circuit I’d found before looking at this thread, I did add the 1 meg "safety resistor" across the bias balance pot to ensure that there's always some bias voltage going to the power tube grid.  Probably overkill, but sluckey & I will sleep better   Of course, you don't need it for the circuit FYL posted - another element of elegance.

Chip

[jjasilli]

do the two caps in Yves' circuit provide just as much filtering as the alternative. It looks that way to me. Eer, I mean, "Oui!"

If you know your raw bias voltage  (AlimNeg) and your target grid voltage (Vg), how do you calculate the value of R2 and R6?  It looks to me that these 2 resistors coukld be just 1, with 2 feeds off of it.  It would be easy to experiment with values with tubes OUT.  Because no current is involved, as Sluckey pointed out, just play around and measure bias voltage on the signal grid pin.

you don't need it for the circuit FYL posted - another element of elegance. The bigger danger there is is in the shunt pot.  If that goes open -- Sacre Bleu!

[Fresh_Start]

The bigger danger there is is in the shunt pot.  If that goes open -- Sacre Bleu!

 

I don't follow.  If the wiper of the shunt pot (variable resistor going to ground) goes out, you have raw bias voltage on the grids.  It'll sound like crap but won't hurt anything the way having zero (0) bias voltage would.

BTW you have to have those two separate resistors (R2 & R6).  Otherwise, there would just be a short across the balance pot.

Cheers,

Chip

[FYL]

One question:  do the two caps in Yves' circuit provide just as much filtering as the alternative.  If so, it IS more elegant.  Lower parts count too...

Local bypasses establishing a set known time constant for bias - you shouldn't rely on what's upstream unless you've designed it. Yves - aka Dissident Audio -  is from the "ceinture et bretelles" school of design and knows his stuff.

There are some parentheses missing in the formula shown for R2 & R6.

Zis is very French: just track back from the published values and the right parenthesis level will appear. Or not.
 

a Mouser reference in German a few days ago!

Ahem, Swedish actually. Or maybe Mouser does some not so neat tricks with their dynamically generated pages.

[jjasilli]

The bigger danger there is is in the shunt pot.  If that goes open -- Sacre Bleu!

 

I don't follow.  If the wiper of the shunt pot (variable resistor going to ground) goes out, you have raw bias voltage on the grids.  It'll sound like crap but won't hurt anything the way having zero (0) bias voltage would.

BTW you have to have those two separate resistors (R2 & R6).  Otherwise, there would just be a short across the balance pot.

Cheers,

Chip

I think if the French shunt pot fails open, the bias circuit will entirely lose it's ground connection and the bias supply voltage will be lost.  Is this wrong?

[Fresh_Start]

One question:  do the two caps in Yves' circuit provide just as much filtering as the alternative.  If so, it IS more elegant.  Lower parts count too...

Local bypasses establishing a set known time constant for bias - you shouldn't rely on what's upstream unless you've designed it. Yves - aka Dissident Audio -  is from the "ceinture et bretelles" school of design and knows his stuff.

There are some parentheses missing in the formula shown for R2 & R6.

Zis is very French: just track back from the published values and the right parenthesis level will appear. Or not.
 

"Not" so far, but I will keep trying!

a Mouser reference in German a few days ago!

Ahem, Swedish actually. Or maybe Mouser does some not so neat tricks with their dynamically generated pages.

Now I am embarrassed!     The Swedish flag in the upper right-hand corner of the page should have been a clue... 

I can tell the difference between French, Spanish, Italian, Portugese and Catalan, but German, Flemish, Swedish and Norwegian not so much.  (Note: I can't speak any of these languages now but was fluent in Spanish at one point and could understand basic Catalan.)

Chip

[PRR]

> I think if the French shunt pot fails open, the bias circuit will entirely lose it's ground connection

Right.

> and the bias supply voltage will be lost.  Is this wrong?

Yes, wrong.

Tug-of-war between Bias and Ground. Cut the rope to Ground. Who wins? Bias! You wanted say -30V, you now have say -45V, which is a lot better than zero V.

[FYL]

Tug-of-war between Bias and Ground. Cut the rope to Ground. Who wins? Bias! You wanted say -30V, you now have say -45V, which is a lot better than zero V.

Yes, any pot failure results in max bias voltage applied to the tubes, thus protecting them.

[jjasilli]

do the two caps in Yves' circuit provide just as much filtering as the alternative. It looks that way to me. Eer, I mean, "Oui!"
I'm not so sure anymore.  It now looks to me like ea side has only one stage of filtering.

[FYL]

It now looks to me like ea side has only one stage of filtering.

Yves' bias circuit should be used with a fully filtered low Z bias supply. The caps are part of local decoupling nodes cutting what can be left of noise, they are small enough as not to modify time constants - big caps here would mean a "slow" setting bias.

Low Z means a circuit derived from it's own iron, windings or taps, not from B+ with a large dropping resistor as commonly seen. It'd work with a high Z supply but fine-tuning voltages could be tedious.

[Fresh_Start]

Ok, French arithmetic sucks.  Would someone please show me how you fill in the values provided for AlimNeg (-55), Vg (-35), and R4 (10,000) to get 17,800 out of the following formula?

(AlimNeg - Vg) / Vg * R4 * 4 - (R4/2)

There are some parentheses missing and I can't come up with 17K8 any way that makes sense to me.  Perhaps this is new French math?  Perhaps my brain is old and I'm missing something really obvious?  

Either way, please help me out here.

Chip

[sluckey]

((AlimNeg - Vg) / Vg * R4 * 4) - (R4/2) = 17857.14285714286

These are basic rules to follow when solving equations. Looking at the original formula, first solve for any values inside parenthesis. Next, perform any multiplications or divisions. Finally, do the additions or subtractions.

[Fresh_Start]

((AlimNeg - Vg) / Vg * R4 * 4) - (R4/2) = 17857.14285714286

These are basic rules to follow when solving equations. Looking at the original formula, first solve for any values inside parenthesis. Next, perform any multiplications or divisions. Finally, do the additions or subtractions.

Believe it or not, I know and remembered the basic rules.  I thought I'd tried the obvious formula but obviously hadn't:

([(AlimNeg - Vg) / Vg] * R4 * 4) - (R4/2) = 17857.14285714286

No, my name isn't Homer but I sure feel like I need a bottle of Duff beer right about now.  Doah!  

Now for JJ's example, I get -27 as the raw bias voltage ("AlimNeg") assuming we're in the middle of the bias adjustment pot.  Vg is -18.  10k works fine for R4.  So:

([(27 - 18) / 18] * 10,000 * 4) - 5,000 = 15,000

Now THAT is just dumb luck!

However, based on FYL's comments above, the bias needs to be filtered before it ever gets to the "AlimNeg" point.  In that case, the uglier circuit from King Tut actually has a lower parts count.

Chip

[sluckey]

I sure feel like I need a bottle of Duff beer right about now.

I think I'll have an Icehouse with you. 

[jjasilli]

I'm about to give up.  I tried feeding 2 separate bias adjust pots; but they interacted; and acted in parallel so as to reduce the range of the voltage swing.  

Next I tried the bias topology from my Stromberg Carlson amps (somewhat similar to FYL's French circuit) - posted below.  I got the grid leak resistior values by trial & error.  With the stacked pair of grid leak resistors, in parallel with the bias circuit, those values yield about 100K from ea signal grid pin to ground.  This matches the stock value.  

Problems:  

1.  There's only about a 2 volt swing from the full range of the 10K balance pot: from about -9.9VDC -to- about -11.8VDC.  Maybe a larger value pot would give a bigger swing?
2.  I'm getting inconsistent voltage readings & drop-off on one side.  Maybe due to the cheapo mini trim pot pot from Radio Shack.
3.  I need about -8 more volts or so (easy enough to fix that)

[sluckey]

Here's a slight variation on FYL's circuit. This came from a working 45 watt Marshall clone. You'll need to juggle some resistor values since you don't use a PT with a bias tap. Points G and H would connect to the grid leak resistors.

I don't think you will get the same bias voltage sweep from the balance pot as you get from the 'bias adjust' pot. If you want to have the same voltage adjust, then just duplicate the entire bias circuit all the way back to the PT HT winding. There will be plenty of adjustment with no interaction.

[jjasilli]

Sluckey, thanks for the moral support!  I stuck with the Stromberg Carlson approach, as it was already built.  The swing of the balance pot IS good enough to get the tubes in balance.  Wish it was a multi-turn pot, but now I can get both tubes within 1mA (measured across Doug's 1Ω 1% 3W bias sense resistors).  Values are drifting a bit, as I have a new B+ filter caps burning-in.  There's proper current draw too, but at the far end of the adjust pot.  I'll have to play with the value of the shunt resistor (now 68K) to get everything in balance. 

Any ideas on filtering the bias circuit?  Some circuits use one 50uF cap.  Doug's improved circuit uses 2X 10uF caps straddling the dropping resistor to to the bias adjust pot.  My circuit, following Stromberg Carlson, now has 2X 10uF caps as shown in the schematic attached:  one cap is after the dropping resistor; one cap after the adjust pot.  Any advantage to adding a 3rd 10uF cap (or other value) before the dropping resistor?

[Fresh_Start]

I'm no expert, but I might move the 10uf cap now tied to the adjustment pot wiper over to before the 15K resistor (where you show an optional 3rd cap).  That would form a consistent pi filter instead of having the resistance between the two caps varying as you adjust the bias.  However, the French style also included filter cap closer to the grid return resistors assuming the supply was already filtered.

I did play with Duncan's PSU a while ago comparing one big cap to 2 small caps separated by a resistor and Hoffman's pi filter with two 10uf caps and a 15K resistor was much more effective at reducing ripple than a 100uf cap by itself.

Maybe someone who actually knows how to design electrical circuits can chime in.

Cheers,

Chip

[sluckey]

I'd put a 10µF on each side of that 15K to form an RC pi filter. Then I'd drop the cap on the wiper of the pot. And finally, put a 10µF across each 47K resistor. Total of 4 caps.

[jjasilli]

Finally got back to this.  Seems my power tubes are way mismatched; couldn't get the bias balance pot to balance the bias!   
Well not really that mad.  So I ripped out the balance circuit and wired-up dual bias adj pots.  To my surprise they're amazingly interactive! 

The stock amp has 400VDC plate volts and very cold bias.  I installed a 12V bucking transformer feeding 109VAC into the PT primary.  There's a 330Ω power resistor in the B+ dropping about 20VDC.  With more current draw for better bias, there's now 285 plate volts drawing 38mA for 10.83W plate diss.  A tad hot but close enough.  The amp never sounded better, the brittleness, caused by over-voltage is gone from the el-84's!